1. **Problem statement:** We have a normally distributed variable (years of schooling) with standard deviation $\sigma = 3.5$ years. A confidence interval (CI) for the mean is given as $(13.8, 15.9)$ based on a sample size $n=64$. We want to find: a. The confidence level $1-\alpha$. b. The minimum sample size to reduce the margin of error (ME) to 0.9 years. c. The minimum sample size to reduce the length of the CI by 20%. 2. **Formulas and rules:** - The confidence interval for the mean when $\sigma$ is known is: $$\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$ - Margin of error (ME) is: $$ME = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$ - Length of CI is $2 \times ME$. - To find $z_{\alpha/2}$, use the margin of error and known $\sigma, n$. - To find sample size $n$ for a desired ME: $$n = \left(\frac{z_{\alpha/2} \sigma}{ME}\right)^2$$ 3. **Step a: Find confidence level** - Sample mean $\bar{x} = \frac{13.8 + 15.9}{2} = 14.85$ - Margin of error $ME = 15.9 - 14.85 = 1.05$ - Using $ME = z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$: $$z_{\alpha/2} = \frac{ME \sqrt{n}}{\sigma} = \frac{1.05 \times 8}{3.5} = \frac{8.4}{3.5} = 2.4$$ - Find confidence level from $z_{\alpha/2} = 2.4$: The two-tailed confidence level is $1 - \alpha = 2 \Phi(2.4) - 1$ where $\Phi$ is the standard normal CDF. - Using standard normal tables or calculator, $\Phi(2.4) \approx 0.9918$. - So confidence level: $$1 - \alpha = 2 \times 0.9918 - 1 = 0.9836 = 98.36\%$$ 4. **Step b: Minimum sample size for margin of error 0.9** - Use formula: $$n = \left(\frac{z_{\alpha/2} \sigma}{ME}\right)^2 = \left(\frac{2.4 \times 3.5}{0.9}\right)^2$$ - Calculate numerator: $$2.4 \times 3.5 = 8.4$$ - Divide by ME: $$\frac{8.4}{0.9} = 9.3333$$ - Square: $$9.3333^2 = 87.11$$ - Minimum sample size: $$n = 88$$ (round up to nearest whole number) 5. **Step c: Minimum sample size to reduce length by 20%** - Original length $L = 2 \times ME = 2 \times 1.05 = 2.1$ - New length $L_{new} = 0.8 \times 2.1 = 1.68$ - New margin of error $ME_{new} = \frac{L_{new}}{2} = 0.84$ - Use formula for $n$: $$n = \left(\frac{z_{\alpha/2} \sigma}{ME_{new}}\right)^2 = \left(\frac{2.4 \times 3.5}{0.84}\right)^2$$ - Calculate numerator: $$2.4 \times 3.5 = 8.4$$ - Divide by new ME: $$\frac{8.4}{0.84} = 10$$ - Square: $$10^2 = 100$$ - Minimum sample size: $$n = 100$$ **Final answers:** - a. Confidence level is approximately 98.36%. - b. Minimum sample size for ME = 0.9 is 88. - c. Minimum sample size to reduce length by 20% is 100.